$ \int_0^4 \int_0^{y^2} dx \, dy$ Switch the bounds of the double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^{16} \int_{\sqrt{x}}^4 dy \, dx$ (Choice B) B $ \int_0^{16} \int_0^{x^2} dy \, dx$ (Choice C) C $ \int_0^4 \int_0^{16} dy \, dx$ (Choice D) D $ \int_0^{16} \int_0^{\sqrt{x}} dy \, dx$
Explanation: The first step whenever we want to switch bounds is to sketch the region of integration that we're given. Here, we see $0 < x < y^2$ and $0 < y < 4$. Therefore: ${2}$ ${4}$ ${6}$ ${8}$ ${10}$ ${12}$ ${14}$ ${16}$ ${2}$ ${4}$ ${6}$ ${\llap{-}2}$ $y$ $x$ Because we're switching bounds to $dy \, dx$, we need to start with numeric bounds for $x$. We see that $0 < x < 16$. Then we can define $y$ in terms of $x$. Thus, $\sqrt{x} < y < 4$. In conclusion, the double integral after switching bounds is: $ \int_0^{16} \int_{\sqrt{x}}^4 dy \, dx$